If $m>n>0$ and $\frac{1}{m} + \frac{1}{n}$ = $\frac{1}{9}$, find all possible values for $(m, n)$

Question

From a problem in a recent math competition. Was wondering if it was possible to logically solve for the answer instead of guessing and checking.

If $m>n>0$ and $\frac{1}{m} + \frac{1}{n}$ = $\frac{1}{9}$, find all possible values for $(m, n)$

Answer 1

The expression $\frac{1}{m}+\frac{1}{n}=\frac{1}{9}$ is equivalent to $$81=(m-9)(n-9)$$

The divisors of $81$ are $\pm1,\pm3,\pm9,\pm27,\pm81$.

Answer 2

Rewrite the relation as $9m + 9n = mn$.

Solving, we see that the only solutions (say over the positive reals) are of the form $(9n/(n-9), n)$.

Clearly we must have $n > 9$. Additionally, the constraint $m > n$ implies $9/(n-9) > 1$. So $n < 18$.

Now you can just check $n\in \{10, 11, 12, 13, 14, 15, 16, 17\}$, to see when $9n/(n-9)$ is an integer.

The solutions are thus with $n \in \{10, 12\}$, and plugging into $m = 9n/(n-9)$, the pairs are $$ (90, 10), \quad \mbox{and} \quad (36, 12). $$

Answer 3

I'm going to assume $m$ and $n$ are integers.

Since $\frac{1}{m}=\frac{1}{9}-\frac{1}{n}$ and $m,n>0$, we know that $n\ge 10$. Also, since $\frac{1}{9}-\frac{1}{n}=\frac{n-9}{9n}=\frac{1}{m}$, we know that $m=\frac{9n}{n-9}$. So $m>n$ gives us $\frac{9n}{n-9}>n$ which simplifies to $0>n(n-18)$ and since $0<n$ this means $0>n-18$, so we now have $10 \le n < 18$.

Thus $n=10, 11, \dots, 17$ and plugging in these values to $\frac{9n}{n-9}$ should give you the corresponding $m$-values.

Answer 4

I presume that $m$ and $n$ are integers: otherwise there would be an infinity of answers.

Clearly neither integer can be less than $10$, since that would make the other negative. We may write the equation as $$(m-9)(n-9)=81.$$ Because $81=3^4$, it can factorize exactly three ways into two factors: $81=1\times81=3\times27=9\times9$. Adding $9$ to these factors gives the complete set of solutions: $$(m,n)\in\{(90,10),(36,12)\},$$ which comprises the only two solutions remaining after applying the condition $m>n$.